How does one simplify this boolean expression?

Question

(a + b)(b' + c')(a + b' + c)

where b' = b not and c' = c not.

I tried distributive because I'm not very good at applying the properties when multiplication is applied but I can with addition.

(a + b)(b' + c') = ab' + ac' + b'b + bc' = ab' + ac' + bc' I think it can be simplified further but I'm not sure how. It would just result in two different terms in the distributive property no matter how you look at it, and combining it with the other equation via distributive would just create a mess.

Am I missing something?

Answer 1

Yes your guess that it can be simplified further is right, but not so easy to see.

$ab'+ac'+bc' = ab'+bc'$ because the middle term is covered in all cases ($b$ or $b'$) by one of the other two.

That said, it's still a slight chore to multiply $(ab'+bc')(a+b'+c)$. I don't see an easy way out of this. In general, beyond a few variables it does get very difficult to minimize boolean expressions.

Answer 2

Recall that each operation distributes over the other, so that $(x+y)(x+z)=x+yz$. This suggests that expanding $(a+b)(a+b'+c)$ might be a good approach, since part of the expansion will be the product $(a+b)(a+b')$. And in fact we find that

$$\begin{align*} (a+b)(a+b'+c)&=(a+b)(a+b')+(a+b)c\\ &=a+bb'+(a+b)c\\ &=a+0+ac+bc\\ &=a(1+c)+bc\\ &=a+bc\;. \end{align*}$$

Multiplying this by $b'+c'$ now yields

$$\begin{align*} (a+bc)(b'+c')&=ab'+ac'+bcb'+bcc'\\ &=ab'+ac'\;, \end{align*}$$

which can be checked with a truth table.